0.5g of organic compound in Kjeldahl's method liberated ammonia, which neutralised 60 ml of 0.1 N H2SO4 solution. The percentage of nitrogen in the compound is

% of N= No.of m. eq. of NH3 liberated ×1.4 weight of organic compound 60×0.1×1.40.5=16.8%

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0.5g of organic compound in Kjeldahl's method liberated ammonia, which neutralised 60 ml of 0.1 N H₂SO₄ solution. The percentage of nitrogen in the compound is

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answer is 16.8.

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Detailed Solution

$\begin{array}{l} % of N = \frac{No.of m . eq. of {NH}_{3} liberated \times 1.4}{weight of organic compound} \\ \frac{60 \times 0.1 \times 1.4}{0.5} = 16.8 % \end{array}$

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